Some properties for \(\alpha\)-starlike functions with respect to \(k\)-symmetric points of complex order

Lemma 1. ([10]). Let $\mathbb{k}$ , $\vartheta$ be complex numbers. Suppose that h(z) is convex and univalent in $\triangle$ with (1.8) $$h(0) = 1 \quad and \quad \text{Re}\left[\mathbb{k}h(z) + \vartheta\right] > 0 \quad (z \in \triangle),$$ and let q(z) be analytic in $\triangle$ with q(0) = 1 and $q(z) \prec h(z)$ . If $p(z) = 1 + p_1 z + p_2 z^2 + \cdots \in \wp$ with p(0) = 1, then $$p(z) + \frac{zp'(z)}{kq(z) + \vartheta} \prec h(z)$$ implies that $p(z) \prec h(z)$ .

Lemma 2. (see [6, 7]). Let $\mathbb{k}$ , $\vartheta$ be complex numbers. Suppose that h(z) is convex and univalent in $\triangle$ and satisfies (1.8). If $p(z) = 1 + p_1 z + p_2 z^2 + \ldots$ $\in \wp$ and satisfies the subordination $$p(z) + \frac{zp'(z)}{\mathbb{k}p(z) + \vartheta} \prec h(z),$$ then $$p(z) \prec h(z)$$ . 2. Main result. Unless otherwise mentioned, we assume throughout this article that $f \in \mathcal{A}$ , $\alpha > 0$ , $\varphi \in \wp$ and $\gamma \in \mathbb{C}^*$ .

Proposition 1. Let $f \in \mathcal{T}_{\gamma,\alpha}(\varphi)$ and $\operatorname{Re} \frac{1}{\alpha} [\alpha \gamma (\varphi(z) - 1) + 1] > 0$ , then $f \in \mathcal{S}_{\gamma}(\varphi)$ . Proof. Set (2.1) $$p(z) = 1 + \frac{1}{\gamma} \left( \frac{zf'(z)}{f(z)} - 1 \right),$$ then p is an analytic function with p(0) = 1. By differentiating (2.1) logarithmically, we get $$1 + \frac{1}{\gamma} \left( \frac{\alpha z (zf'(z))' + (1 - \alpha)zf'(z)}{\alpha zf'(z) + (1 - \alpha)f(z)} - 1 \right)$$ $$= p(z) + \frac{\alpha zp'(z)}{1 + \alpha\gamma [p(z) - 1]} \prec \varphi(z), \quad (z \in \triangle).$$ The conclusion of Proposition 1 yields from Lemma 2, by taking $\mathbb{k} = \gamma$ and $\vartheta = \frac{1-\alpha\gamma}{\alpha}$ . Similarly, we can prove the following proposition.

Proposition 2. Let Re $\frac{1}{\alpha} \left[ \alpha \gamma \left( \varphi(z) - 1 \right) + 1 \right] > 0$ . Then (2.2) $$F(z) = I_{\alpha}(f) = \frac{1}{\alpha z^{(1/\alpha)-1}} \int_{0}^{z} t^{(1/\alpha)-2} f(t) dt \in \mathcal{S}_{\gamma}(\varphi)$$ whenever $f(z) \in \mathcal{S}_{\gamma}(\varphi)$ .

Theorem 1. Let $f \in \mathcal{T}^k_{\gamma,\alpha}(\varphi)$ and $\operatorname{Re} \frac{1}{\alpha} [1 + \alpha \gamma (\varphi(z) - 1)] > 0$ . Then $f_k$ defined by (1.2) is in $\mathcal{T}_{\gamma,\alpha}(\varphi)$ . Further, we have $f_k \in \mathcal{S}_{\gamma}(\varphi)$ . Proof. Let $f \in \mathcal{T}^k_{\gamma,\alpha}(\varphi)$ . Replacing z by $\varepsilon^{\mu}z$ ( $\mu = 0, 1, \ldots, k-1$ ; $\varepsilon^k = 1$ ) in (1.4), then (1.4) also holds true, that is, $$(2.3) 1 + \frac{1}{\gamma} \left( \frac{(1-\alpha)\varepsilon^{\mu}zf'(\varepsilon^{\mu}z) + \alpha\varepsilon^{\mu}z\left[f'(\varepsilon^{\mu}z) + \varepsilon^{\mu}zf''(\varepsilon^{\mu}z)\right]}{(1-\alpha)f_{k}(\varepsilon^{\mu}z) + \alpha\varepsilon^{\mu}zf'_{k}(\varepsilon^{\mu}z)} - 1 \right) \prec \varphi(z).$$ According to the definition of $f_k$ and $\varepsilon^k = 1$ , we know that (2.4) $$f_k(\varepsilon^{\mu}z) = \varepsilon^{\mu} f_k(z) \text{ and } f'_k(\varepsilon^{\mu}z) = f'_k(z) = \frac{1}{k} \sum_{\mu=0}^{k-1} f'(\varepsilon^{\mu}z),$$ for any $\mu = 0, 1, \dots, k-1$ , and summing up, we can get $$\frac{1}{k} \sum_{\mu=0}^{k-1} \left[ 1 + \frac{1}{\gamma} \left( \frac{(1-\alpha)zf'(\varepsilon^{\mu}z) + \alpha z \left[ zf'(\varepsilon^{\mu}z) \right]'}{(1-\alpha)f_k(z) + \alpha z f'_k(\varepsilon^{\mu}z)} - 1 \right) \right] \\ = 1 + \frac{1}{\gamma} \left( \frac{(1-\alpha)zf'_k(z) + \alpha z \left[ zf'_k(z) \right]'}{(1-\alpha)f_k(z) + \alpha z f'_k(z)} - 1 \right).$$ Hence there exist $\zeta_{\mu}$ 's in $\triangle$ such that $$1 + \frac{1}{\gamma} \left( \frac{(1 - \alpha)z f_k'(z) + \alpha z \left[ z f_k'(z) \right]'}{(1 - \alpha)f_k(z) + \alpha z f_k'(z)} - 1 \right) \prec \frac{1}{k} \sum_{\mu=0}^{k-1} \varphi(\zeta_{\mu}) = \varphi(\zeta_0),$$ for $\zeta_0 \in \Delta$ , since $\varphi(\Delta)$ is convex. Thus $f_k \in \mathcal{T}_{\gamma,\alpha}(\varphi)$ . Remark 4. Putting $\gamma = 1 - \lambda$ $(0 \le \lambda < 1)$ and $\varphi(z) = \frac{1+z}{1-z}$ in the above theorem, we get the result obtained by Wang et al. [19, Lemma 3, p. 110].

Theorem 2. Let $f \in \mathcal{T}_{\gamma,\alpha}^k(\varphi)$ and $\operatorname{Re} \frac{1}{\alpha} [1 + \alpha \gamma [\varphi(z) - 1]] > 0$ . Then $f \in \mathcal{S}_{\gamma}^k(\varphi)$ . Proof. Let $f \in \mathcal{T}^k_{\gamma,\alpha}(\varphi)$ . Then by Definition 1, we have $$1 + \frac{1}{\gamma} \left( \frac{\alpha z \left( z f'(z) \right)' + (1 - \alpha) z f'(z)}{\alpha z f'_k(z) + (1 - \alpha) f_k(z)} - 1 \right) \prec \varphi(z).$$ Putting $p(z) = 1 + \frac{1}{\gamma} \left( \frac{zf'(z)}{f_k(z)} \right) - 1$ and $q(z) = 1 + \frac{1}{\gamma} \left( \frac{zf'_k(z)}{f_k(z)} \right) - 1$ , it is easy to obtain that $$1 + \frac{1}{\gamma} \left( \frac{\alpha z \left( z f'(z) \right)' + (1 - \alpha) z f'(z)}{\alpha z f'_k(z) + (1 - \alpha) f_k(z)} - 1 \right)$$ $$= p(z) + \frac{\alpha z p'(z)}{1 + \alpha \gamma \left[ q(z) - 1 \right]}$$ Since $f \in \mathcal{T}_{\gamma,\alpha}^k(\varphi)$ , then by using Theorem 1, we can see that $q(z) \prec \varphi(z)$ . Now an application of Lemma 1, yields $$p(z) = 1 + \frac{1}{\gamma} \left( \frac{zf'(z)}{f_k(z)} - 1 \right) \prec \varphi(z).$$ That is $f \in \mathcal{S}_{\gamma}^{k}(\varphi)$ . We thus complete the proof of Theorem 2.

Theorem 3. Let $f \in \mathcal{S}_{\gamma}^{k}(\varphi)$ and $\operatorname{Re} \frac{1}{\alpha} [1 + \alpha \gamma (\varphi(z) - 1)] > 0$ in $\triangle$ and let F be the integral operator defined by (2.2), then $F \in \mathcal{S}_{\gamma}^{k}(\varphi)$ . Proof. Let a function $f_k(z)$ of the form (1.2) with F(z) be put in the place of f(z). That is $F_k(z) = \frac{1}{k} \sum_{v=0}^{k-1} \varepsilon^{-v} f(\varepsilon^v z)$ . We can see that $F_k(z) = \frac{1}{\alpha z^{(1/\alpha)-1}} \int_0^z \left(t^{(1/\alpha)-2} f_k(t)\right) dt$ and then differentiating with respect to z, we get $$(2.5) (1-\alpha)F_k(z) + \alpha z F'_k(z) = f_k(z).$$ From (2.2), we have (2.6) $$(1 - \alpha)F(z) + \alpha z F'(z) = f(z).$$ Since $f \in \mathcal{S}_{\gamma}^{k}(\varphi)$ , we can apply Theorem 1 with $\alpha = 0$ to deduce $f_{k} \in \mathcal{S}_{\gamma}(\varphi)$ . Applying Proposition 2, we have $F_{k} \in \mathcal{S}_{\gamma}(\varphi)$ , that is (2.7) $$1 + \frac{1}{\gamma} \left( \frac{z F_k'(z)}{F_k(z)} - 1 \right) \prec \varphi(z).$$ If we denote (2.8) $$p(z) = 1 + \frac{1}{\gamma} \left( \frac{zF'(z)}{F_k(z)} - 1 \right),$$ and (2.9) $$q(z) = 1 + \frac{1}{\gamma} \left( \frac{z F_k'(z)}{F_k(z)} - 1 \right),$$ then p(z) is analytic in $\triangle$ with p(0) = 1 and q(z) is analytic in $\triangle$ with q(0) = 1, $q(z) \prec \varphi(z)$ . Differentiating in (2.6) and using (2.8), we have $$(2.10) \gamma p(z) - \gamma + 1 + \frac{\alpha \gamma z p'(z)}{\left(1 - \alpha + \alpha \frac{z F_k'(z)}{F_k(z)}\right)} = \frac{z f'(z)}{(1 - \alpha) F_k(z) + \alpha z F_k'(z)}.$$ In view of (2.5) and (2.9), (2.10) gives $$p(z) + \frac{\alpha z p'(z)}{1 + \alpha \gamma \left(q(z) - 1\right)} = 1 + \frac{1}{\gamma} \left(\frac{z f'(z)}{f_k(z)} - 1\right) \prec \varphi(z).$$ Now applying Lemma 1, we get $p(z) \prec \varphi(z)$ , which proves the theorem. The method of proving Theorem 4 is similar to that of Theorem 2.

Theorem 4. Let Re $$\frac{1}{\alpha} [1 + \alpha \gamma [\varphi(z) - 1]] > 0$$ . Then $\mathcal{F}_{\gamma,\alpha}^k(\varphi) \subset \mathcal{F}_{\gamma,0}^k(\varphi)$ . Now, we give the integral representations of functions belonging to the classes $\mathcal{T}_{\gamma,\alpha}^k(\varphi)$ .

Theorem 5. Let $f \in \mathcal{T}^k_{\gamma,\alpha}(\varphi)$ , then we have (2.11) $$f_k(z) = \frac{1}{\alpha} z^{1-\frac{1}{\alpha}} \int_0^z \exp\left\{ \frac{\gamma}{k} \sum_{\mu=0}^{k-1} \int_0^{\varepsilon^{\mu} t} \frac{\varphi(\omega(\xi)) - 1}{\xi} d\xi \right\} t^{\frac{1}{\alpha} - 1} dt,$$ where $f_k(z)$ is given by equality (1.2) and $\omega(z) \in \Omega$ . Proof. Suppose that $f \in \mathcal{T}_{\gamma,\alpha}^k(\varphi)$ . We know that the condition (1.5) can be written as follows (2.12) $$1 + \frac{1}{\gamma} \left( \frac{\alpha z (zf'(z))' + (1 - \alpha)zf'(z)}{\alpha z f'_k(z) + (1 - \alpha)f_k(z)} - 1 \right) = \varphi(\omega(z)).$$ By similar application of the arguments given in the proof for Theorem 1 to (2.12), we obtain $$(2.13) 1 + \frac{1}{\gamma} \left( \frac{\alpha z \left( z f_k'(z) \right)' + (1 - \alpha) z f_k'(z)}{\alpha z f_k'(z) + (1 - \alpha) f_k(z)} - 1 \right) = \frac{1}{k} \sum_{\mu=0}^{k-1} \varphi(\omega(\varepsilon^{\mu} z)).$$ From (2.13), we have (2.14) $$\frac{\alpha (zf'_k(z))' + (1-\alpha)f'_k(z)}{\alpha zf'_k(z) + (1-\alpha)f_k(z)} - \frac{1}{z} = \frac{\gamma}{k} \sum_{\mu=0}^{k-1} \frac{\varphi(\omega(\varepsilon^{\mu}z)) - 1}{z}.$$ Integrating this equality, we get $$\log\left[\frac{\alpha z f_k'(z) + (1-\alpha) f_k(z)}{z}\right] = \frac{\gamma}{k} \sum_{\mu=0}^{k-1} \int_0^z \frac{\varphi(\omega(\varepsilon^{\mu}\rho)) - 1}{\rho} d\rho,$$ that is, (2.15) $$\alpha z f_k'(z) + (1 - \alpha) f_k(z) = z \exp \frac{\gamma}{k} \sum_{\mu=0}^{k-1} \int_0^z \frac{\varphi(\omega(\varepsilon^{\mu}\rho)) - 1}{\rho} d\rho,$$ or equivalently, $$(2.16) \qquad (1-\alpha) f_k(z) + \alpha z f_k'(z) = z \exp \frac{\gamma}{k} \sum_{\mu=0}^{k-1} \int_0^{\varepsilon^{\mu} z} \frac{\varphi(\omega(\zeta)) - 1}{\zeta} d\zeta.$$ From (2.16), we can get equality (2.11) easily. This completes the proof of Theorem 5.

Definition 1. Let $\mathcal{T}_{\gamma,\alpha}^k(\varphi)$ denote the class of functions f in $\mathcal{A}$ satisfying the following condition $$(1.4) 1 + \frac{1}{\gamma} \left( \frac{\alpha z \left( z f'(z) \right)' + (1 - \alpha) z f'(z)}{\alpha z f'_k(z) + (1 - \alpha) f_k(z)} - 1 \right) \prec \varphi(z),$$ where $\varphi(z) \in \wp$ and $\alpha \geq 0$ .

Definition 2. Let $\mathcal{F}_{\gamma,\alpha}^k(\varphi)$ denote the class of functions $f \in \mathcal{A}$ satisfying the subordination condition $$(1.5) 1 + \frac{1}{\gamma} \left( \frac{\alpha z \left( z f'(z) \right)' + (1 - \alpha) z f'(z)}{\alpha z \xi'_k(z) + (1 - \alpha) \xi_k(z)} - 1 \right) \prec \varphi(z),$$ where $$\xi_k(z) = \frac{1}{k} \sum_{v=0}^{k-1} \varepsilon^{-v} \xi(\varepsilon^v z), \ \xi(z) \in \mathcal{T}_{\gamma,\alpha}^k(\varphi), \ \varphi(z) \in \wp, \ \text{and} \ \alpha \geq 0.$$ By giving specific values to the parameters $k, \gamma$ and $\alpha$ in the class $\mathcal{T}_{\gamma,\alpha}^k(\varphi)$ , we get the following new subclass of analytic functions. Remark 1. Putting k = 1, we obtain the following definition.

Definition 3. Let $\mathcal{T}_{\gamma,\alpha}(\varphi)$ denote the class of functions $f \in \mathcal{A}$ satisfying the following condition (1.6) $$1 + \frac{1}{\gamma} \left( \frac{\alpha z \left( z f'(z) \right)' + (1 - \alpha) z f'(z)}{\alpha z f'(z) + (1 - \alpha) f(z)} - 1 \right) \prec \varphi(z),$$ where $\varphi(z) \in \wp$ and $\alpha \geq 0$ . Remark 2. If we set $\alpha = 0$ , we have the following definition.

Definition 4. Let $\mathcal{S}_{\gamma}^{k}(\varphi)$ denote the class of functions f in $\mathcal{A}$ satisfying the following condition (1.7) $$1 + \frac{1}{\gamma} \left( \frac{zf'(z)}{f_k(z)} - 1 \right) \prec \varphi(z),$$ where $\varphi(z) \in \wp$ .